Hi DPB,

I would argue that the pattern I provided is not rabbit-out-of-the-hat at all. I think it will always be there for a (swordfish based, maybe not some of the extensions like the singles free ride etc?) JE that has a pool of just 3 base candidates. Why? Because the green truths are from the fish that comprise the exocet, the purple truths are from the complementary fish (these must "line up" with the base cells otherwise the base cells wouldn't be "in the way" of the fish and we wouldn't have an exocet) and the cell truths are obviously from the base of the exocet.

The trick is the virtual set. There is no need to panic here. It is just a collection of elements at least one of which must be true, just like a native set, except a native set obviously has exactly one true. In r357c46 it is impossible for example to have the combination 1r3c4, 3r3c6, 6r5c4, 1r5c6, 3r7c4 and 6r7c6. Any combination in those six cells solely from the pool of JE base candidates {1,3,6 in this case} is impossible. Why? Well imagine that such a combination was the solution, and every other cell of the puzzle was filled in. There would be two candidates from {1,3,6} available for r3c46, because one of r3c1379 would have been filled in with a 1,3 or 6, and likewise for r5c357 and r7c357. For definiteness let's suppose that we are left with {1,3} in r3c46, {3,6} in r5c46 and {1,6} in r7c46. How do you solve the puzzle from there? There are two possible solutions: {1r3c4, 3r3c6, 3r5c4, 6r5c6, 6r7c4, 1r7c6} and {3r3c4, 1r3c6, 6r5c4, 3r5c6, 1r7c4, 6r7c6}. This is a contradiction under the assumption that the solution is unique, so it must be that such a combination can't be the solution, therefore one of the "uniqueness spoilers", that is, elements of the virtual set must be true.

Now we're home. We have a structure (pattern) with at least 15 true candidates (from the 15 truths), and at most 15 true candidates (from the 15 links) and no triplets, i.e. exactly 15 true candidates (the classic fish principle). Everything in the pattern is rank 0, and we get the bonus of now knowing that exactly one element of the virtual set is true, i.e. the virtual set is both strong and weak. For the manual solver this pattern should be easy to apply because it only needs to be constructed after a JE with a pool of 3 base digits is found. You don't have to go looking for it. As I said, I think that this pattern will always exist for a classic JE with a pool of just 3 base candidates, as everything needed has to be there!

Here it is applied to PlatinumBlonde (one extra truth and link because it is a twin, sorry JE3 in the new nomenclature):

54 Candidates,

16 Truths = {1V1 679R4 4679R7 679C3 679C4 12N7}

16 Links = {679c7 12n3 12n4 47n8 7n9 6b57 7b48 9b48}

13 Eliminations --> r4c8<>234, r6c57<>6, r6c17<>7, r9c57<>9, r7c8<>23, r7c9<>1, r8c6<>7,

Finally, I must confess that I'm not fully across your incompatible base pair heuristics. To me the technique seems to be about what candidates are remaining in a box that might form URs, but I'm not getting the link between that and which band certain candidates lie in. I'm sure something will click on my next pass through your description, which I'm off to right now. Once I understand that then I'll be able to follow your start for FM.

P.S. I like to use Xsudo to find/prove new patterns which I can then use to solve puzzles. Remember that the exocet pattern was found as a rabbit-out-of-the-hat elimination in Xsudo, and it then took a while to understand what was really going on.

SV